Mehedi33's blog

By Mehedi33, 5 years ago, In English,
 
 
 
 
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5 years ago, # |
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the input size if small (the largest factorial you will need is 20!) so here you can generate all the possible factorials and use bitmasking (or recursion) to pre-calculate all possible values and cache them ... then for every case you just output the answer

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    5 years ago, # ^ |
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    I pre-calculate all the value upto 20. But now i cant understand how can i find my desired number .

    I have no idea about bitmasking. Is there any other way ?

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      5 years ago, # ^ |
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      Use brute-force. For each number there are two cases — to use or not to use :)

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        5 years ago, # ^ |
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        If i use brute-force, at some stage summation of fact(num) may exceed my desired n , at that case i may avoid that fact(num) . But can i surely say that at any stage summation of fact(num) will be equal to my desired n ?

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          5 years ago, # ^ |
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          I think that this approach will result in TLE since the limit is 0.5 :) I think it is better to read Caraz96's idea and ask if you misunderstood something in it.

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      5 years ago, # ^ |
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      It is possible to use a greedy approach: since fact(n) > fact(0)+fact(1)+...+fact(n-1) (this is true for each n > 2) you can iterate from the greatest factorial(20!) to the smallest one, if fact[i] is less than or equal to N you must take it, decrease N by fact[i] and proceed with the next factorial. At the end, if N is 0 you found the solution, otherwise it is impossible to obtain N adding factorials.

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    3 years ago, # ^ |
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    Exactly I started it today and I went totally in this way . As a very beginner I was amazed when I saw that other coders also think in the same way .